Eduardo Kausel-Fundamental solutions in elastodynamics_ a compendium-Cambridge University Press (2006)

110 Solution to the Helmholtz and wave equations
and since Ψ 1 must satisfy ∇×∇×Ψ 1 = kS 2 Ψ 1, then
( )
[
∂ψ
∇ − ˆk ∇ 2 ψ = kS
2 ψ ˆk + 1 ( )] ∂ψ
∂z
kS 2 ∇
∂z
(8.115)
which results in the condition ∇ 2 ψ + kS 2 ψ = 0. Similarly, the second term gives
( ) ∂χ
k S ∇×Ψ 2 =∇ − ˆk ∇ 2 χ (8.116)
∂z
so
( ) ∂χ
)
k S ∇×∇×Ψ 2 =∇×∇ −∇×(
ˆk ∇ 2 χ = kS 3 ∂z
Ψ 2
) ( )
=−∇×(
ˆk ∇ 2 χ = kS 2 ∇× ˆk χ
(8.117)
∇×[
ˆk ( ∇ 2 χ + k 2 S χ)] = 0 (8.118)
which implies ∇ 2 χ + kS 2 χ = 0. Thus, the solution provided does indeed satisfy the vector
Helmholtz equation. It remains to verify the gauge condition:
( )
∇ · Ψ =∇· ψ ˆk + 1 ( ) ∂ψ
kS 2 ∇ · ∇ + 1 ( )
∇ · ∇× χ ˆk
∂z k S
( )
=∇· ψ ˆk + 1 [
kS 2 ∇ · ∇×∇×
( ) ]
ψ ˆk + ˆk ∇ 2 ψ
( ) ( )
=∇· ψ ˆk −∇· ψ ˆk = 0 (8.119)
Hence, the solution to the vector Helmholtz equation can now be formed with the solutions
to the two scalar Helmholtz equations for ψ and χ, using for this purpose the expressions
provided in the previous section:
ψ(r,θ,z) = (c 1 cos nθ + c 2 sin nθ)(c 3 J n (k β r) + c 4 Y n (k β r)) ( c 5 e ikzz + c 6 e −ikzz) (8.120)
χ(r,θ,z) = (d 1 cos nθ + d 2 sin nθ)(d 3 J n (k β r) + d 4 Y n (k β r)) ( d 5 e ikzz + d 6 e −ikzz) (8.121)
in which the c i , d i are arbitrary constants, and
√
k β = kS 2 − k2 z (8.122)
8.7 Elastic wave equation in cylindrical coordinates
Consider the elastic wave equation for an isotropic medium
(λ + 2µ) ∇∇ · u − µ∇ ×∇×u = ρü (8.123)
Defining
ε =∇· u = volumetric strain (8.124)
Ω = 1 ∇×u = rotation vector, which satisfies ∇ · Ω = 0 (8.125)
2
we can write the elastic wave equation as
(λ + 2µ) ∇ε − 2µ∇ ×Ω = ρü (8.126)